Question

What is len's formula? Derive the formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ for spherical lens.

Answer 1

Lens formula: An equation which relates the object distance, image distance and focal length of a lens is called the lens formula. It is given as

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

The figure shows the formation of a real, inverted and diminished image $AB$ of the object $AB$ placed beyond the centre of curvature at a distance $u$ from the convex lens.

Let v be the image distance.

According to Cartesian sign convention,

Object distance $\left(OB\right)=-u$

Image distance $\left(O{B}^{\prime }\right)=+v$

Focal length $(O{F}_1=O{F}_2)=+f$

From the geometry of the figure above, right angled $△ABO$ and $△A^{\prime }{B}^{\prime }O$ are similar.

$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{OB}{O{B}^{\prime }}=\frac{-u}{v}....\left(1\right)$

Similarly, from the geometry of the figure, right angled $△OD{F}_2$ and $△B^{\prime }{A}^{\prime }{F}_2$ are similar.

$\therefore \frac{OD}{A^{\prime }{B}^{\prime }}=\frac{O{F}_2}{F_2{B}^{\prime }}$

$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{O{F}_2}{F_2{B}^{\prime }}(\because OD=AB)$

$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{O{F}_2}{O{B}^{\prime }-O{F}_2}$

$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{f}{v-f}....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$-\frac{u}{v}=\frac{f}{v-f}$

$\Rightarrow -u(v-f)=vf$

$\Rightarrow -uv+uf=vf$

Dividing each term by $uvf$,

$-\frac{uv}{uvf}+\frac{uf}{uvf}=\frac{vf}{uvf}$

$\therefore -\frac{1}{f}+\frac{1}{v}=\frac{1}{u}$

$\therefore \frac{1}{v}-\frac{1}{v}=\frac{1}{f}$

This equation is the 'Lens formula'.